$\frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{A x+B}{x^2+a x+2}+\frac{C x+D}{x^2+b x+2}$ ...(i)
$\because \frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{x^2-7 x+2}{\left(x^2+x+2\right)\left(x^2-x+2\right)}$
The form of the partial fraction decomposition is :$\frac{x^2-7 x+2}{\left(-x^2-x+2\right)\left(x^2+x+2\right)}=\frac{P x+Q}{x^2-x+2}+\frac{R x+S}{x^2+x+2}$
...(ii)
$\begin{aligned} & \Rightarrow x^2-7 x+2=(P x+Q)\left(x^2+x+2\right)+(R x+S) \\ & \left(x^2-x+2\right) \\ & \Rightarrow x^2-7 x+2=x^3(P+R)+x^2(-P+Q+R+S)+x(2 P \\ & -Q+2 R+S)+(2 Q+2 S)\end{aligned}$
So, by comparing both sides, we get:
$\begin{aligned} & P+R=0 \\ & -P+Q+R+S=1\end{aligned}$
$2 P-Q+2 R+S=-7$
$2 \mathrm{Q}+2 \mathrm{~S}=2$ ...(iii)
Solving system (iii), we get:
$P=0, Q=4, R=0 S=-3$
$\therefore \frac{x^2-7 x+2}{x^4+3 x^2+4}=\frac{4}{x^2+x+2}+\frac{-3}{x^2-x+2}$ ...(iv)
From $\mathrm{eq}^{\mathrm{n}}(\mathrm{i}) \&$ (iv):
$\begin{aligned} & A=0, B=4, C=0, D=-3 \\ & a=1, b=-1 \\ & B+D=4-3=1 \\ & 2 a+b=2-1=1\end{aligned}$
$\therefore B+D=2 a+b$.